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Carnegie Mellon University

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Problem 89 Hard Difficulty

A particle starts from the origin at $t=0$ with a velocity of $8.0 \hat{\mathrm{j}}+2.0 \hat{\mathrm{j}}$ moves in the $x y$ plane with constant acceleration
$(4.0 \hat{\mathrm{i}}+2.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2}$ . When the particle's $x$ coordinate is $29 \mathrm{m},$ what
are its (a) $y$ coordinate and (b) speed?

Answer

$$=45 \mathrm{m}$$
$$=22 \mathrm{m} / \mathrm{s}$$

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Video Transcript

so for party. Given that the initial velocity vector the initial equaling 8.0 meters per second J hat, we can say that the acceleration vector a is giving us 4.0 meters per second squared. I had a plus 2.0 meters per second J hat than the position vector of the particle is equaling the initial velocity vector multiplied by t plus 1/2 times the acceleration vector times t squared. And so this is equaling 8.0 j hat T plus 1/2 multiplied by 4.0 i hat plus 2.0 j hat multiplied by t squared. So we can then say that the position vector is gonna be equaling two 2.0 t squared I hat plus 8.0 t plus 1.0 t squared J hat and so therefore the time that corresponds to ex equaling 29 meters. We can see that solving the equation. We find that 2.0 chief squared is equal ng 29 meters and so t is found to be 3.8 seconds and we're going to solve for the y coordinate at that time. So continuing on for part A. The Y coordinate is gonna be equaling 8.0 meters per second multiplied by 3.8 seconds and this would be plus 1.0 meters per second squared multiplied by 3.8 seconds. Quantity squared. And so this is giving us 45 meters. That would be our answer for part a four car to be. Then the velocity vector of the particle is given as the velocity initial plus the acceleration vector multiplied by time. And so at T equaling again 3.8 seconds. We can say that the velocity vector is gonna be equaling 8.0 meters per second J hat plus 4.0 meters per second squared. I had plus 2.0 meters per second squared J hat multiplied by 3.8 seconds and we find that then the velocity vector at T equals 3.8 seconds is 15 0.2 meters per second. I hat close 15.6 meters per second J hat. So continuing on for part B to find the magnitude of this philosophy, this would be equal in the square root of the X component squared, plus the wide component squared and this is giving us the square root 15.2 meters per second quantity squared plus 15.6 meters per second quantity squared. And we find that then the magnitude of the velocity vector is gonna be giving us 22 meters per second. This would be our magnitude of the velocity vector. AT T equals 3.8 seconds for part B. That is the end of the solution. Thank you for watching.

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